Also , so you should have ,
(add and signs to taste if you don't want to use the
positive root convention).

RonL

Exactly right, and what I've just been scratching out on paper.

I now recommend not looking at the original equation, but the function:
This is the function related to the given equation, so we are looking for the zeros of f(x).

The simplest solution is to graph it. The zeros are obvious...x = [-1,1] as originally advertised. To explain this, note that (as CaptainBlack noted) the convention is to take positive. So, taking a look at f(x):
where we now take the +/- signs to require that the term is positive.

So, on the interval x=(-infinity,-1):
(Since (x+1) is negative on this interval we keep the "-" sign to make it positive...similarly (x-1) is negative and we again keep the "-" sign.)
and thus f(x)=2x+2.

On the interval x=(1,infinity):
and thus f(x)=-2x+2.

Finally, on the interval x=[-1,1]:
and thus f(x)=0.

This is an interesting problem. It gives a strong reminder that we can't always simplify an equation and get the same solution set as we had in the unsimplified equation. It's always a good idea to plug your solution back into the original equation to see if you've got everything. (I'm going to note this problem for when I teach College Algebra again!)

Here is a graph:
(I realized that people have a lot of difficulty when it come to solving absolute value equations, probably has to do with examaning each interval).
Thus the solution set is,
Notice not a finite number of solutions!